Question

1. for the boxes seen below; do the calculations (each environment must equal 100%), draw an arrow to illustrate the direction of water movement. state whether the solution is hypertonic, hypotonic, or isotonic. osmosis water leaves cell. cell shrinks. hypertonic solution 25% h₂o 75% glucose 55% h₂o 45% glucose % h₂o 60% salt 30% h₂o % salt % h₂o 80% oxygen 80% h₂o % oxygen % h₂o 100% solute 95% h₂o % solute 55% h₂o % carbon dioxide 50% h₂o % carbon dioxide 100% h₂o % solute 95% h₂o 3% other % solute % h₂o 88% solute % h₂o 10% other 30% solute 89% h₂o % salt 89% h₂o % other __% salt

Explanation:

First Box (Top - Left, CO₂ and H₂O)

Step1: Calculate % of CO₂

Total is 100%, so \( 100 - 55 = 45 \). So 45% carbon dioxide.

Step2: Inside cell, H₂O is 50%, so solute (CO₂) is \( 100 - 50 = 50 \)%. Compare to outside (45% solute). Since inside solute > outside, water leaves cell? Wait, no: outside H₂O is 55%, inside is 50%. Water moves to lower H₂O (higher solute). So inside has higher solute (50% vs 45% outside), so water leaves cell (hypertonic outside? Wait, no: cell has 50% solute, outside 45% solute. So outside is hypotonic? Wait, maybe I messed up. Wait, the first example: top - right has 25% H₂O outside, 55% inside. So water moves out (arrow right), because outside has less H₂O (more solute: 75% glucose vs 45% inside). So H₂O moves from high H₂O to low H₂O. So for the first left box: outside H₂O 55%, inside H₂O 50%. So H₂O moves from outside (55%) to inside (50%)? Wait, no, the label says "Hypertonic solution: Water leaves cell. Cell shrinks." Wait, maybe the left column is hypertonic solutions (so outside has more solute, less H₂O than cell). So cell has H₂O 50%, so solute 50%. Outside H₂O 55%, so solute 45%. Wait, that would mean outside has less solute (hypotonic), so water would enter cell. But the label says "Hypertonic solution: Water leaves cell". So maybe the left column is cells in hypertonic solutions, so outside has more solute (less H₂O) than cell. So cell H₂O 50%, so outside H₂O should be less than 50%. Wait, the first left box: outside H₂O is 55%? That can't be. Maybe the first left box is a typo, or I misread. Let's do the percentage calculations first.

Top - Right Box (Glucose)

Outside: 25% H₂O, 75% glucose. Inside cell: 55% H₂O, 45% glucose. H₂O moves from inside (55% H₂O) to outside (25% H₂O) (arrow right), because H₂O moves to lower H₂O (higher solute). So outside is hypertonic (more solute, less H₂O) than cell.

Second Right Box (Salt)

Outside: let \( x \) be % H₂O, 60% salt. So \( x + 60 = 100 \) → \( x = 40 \)% H₂O. Inside cell: 30% H₂O, so solute (salt) is \( 100 - 30 = 70 \)%? Wait, no: inside cell: 30% H₂O, so % salt is \( 100 - 30 = 70 \)%. Outside salt is 60%, so outside H₂O is 40%. So H₂O inside is 30%, outside is 40%. So H₂O moves from outside (40% H₂O) to inside (30% H₂O) (arrow left), because H₂O moves to lower H₂O (higher solute). So cell has higher solute (70% vs 60% outside), so outside is hypotonic, water enters cell.

Third Right Box (Oxygen)

Outside: \( x \) % H₂O, 80% oxygen. So \( x = 100 - 80 = 20 \)% H₂O. Inside cell: 80% H₂O, so % oxygen is \( 100 - 80 = 20 \)%. So H₂O inside 80%, outside 20%. So H₂O moves from inside (80% H₂O) to outside (20% H₂O) (arrow right), because outside has less H₂O (more solute: 80% O₂ vs 20% inside). So outside is hypertonic.

Fourth Right Box (Solute)

Outside: \( x \) % H₂O, 100% solute? No, that can't be. Wait, outside: 100% solute? No, total must be 100%. So maybe outside is \( x \) % H₂O, and solute is \( 100 - x \). Inside cell: 95% H₂O, so solute is 5%. So if outside solute is 100%, that's impossible. Maybe it's a typo, and outside is \( x \) % solute, so H₂O is \( 100 - x \). Inside H₂O 95%, solute 5%. So if outside solute is, say, 10% (H₂O 90%), but the problem says "100% Solute" – maybe it's a mistake, and it's 10% solute. Then H₂O outside 90%, inside 95%. So H₂O moves from inside (95%) to outside (90%)? No, inside has more H₂O, so water moves to outside (lower H₂O). But this is confusing. Let's focus on percentage calculations:

First Left Box (CO₂)

Outside: 55% H₂O, so % CO₂ = \( 100 - 55 = 45 \)%. Inside cell: 50% H₂O, so % CO₂ = \( 100 - 50 = 50 \)%.

##…

Answer:

(Selected Calculations):

• Top - Right: Outside H₂O 25%, Inside H₂O 55% (water out, hypertonic outside)
• Second Right: Outside H₂O \( \boldsymbol{40\%} \), Inside Salt \( \boldsymbol{70\%} \) (water in, hypotonic outside)
• Third Right: Outside H₂O \( \boldsymbol{20\%} \), Inside Oxygen \( \boldsymbol{20\%} \) (water out, hypertonic outside)
• First Left: Outside CO₂ \( \boldsymbol{45\%} \), Inside CO₂ \( \boldsymbol{50\%} \)
• Second Left: Outside Solute \( \boldsymbol{0\%} \), Inside Solute \( \boldsymbol{2\%} \)
• Third Left: Inside H₂O \( \boldsymbol{60\%} \), Outside H₂O \( \boldsymbol{12\%} \)
• Fourth Left: Outside Salt \( \boldsymbol{11\%} \) (assuming inside Other + Salt = 11%)

(Note: The problem involves multiple calculations for each box, focusing on percentage composition (total 100%) and water movement direction based on tonicity.)