show your work: (4 points each)\n19) a solution contains 100.0 g water, 10.0 g nacl, and 15.0 g methanol…

show your work: (4 points each)\n19) a solution contains 100.0 g water, 10.0 g nacl, and 15.0 g methanol. what is the weight percent (w/w%) of methanol in the solution?\n\n20) what is the molarity of a solution prepared by dissolving 10.7 g nai in 0.250 l?\n\n21) determine the mass of cacl₂ (mm = 110.98) in 1.75 l of 1.50 m solution\n\n22) a 90.0 g sample of naoh is dissolved in water and the solution is diluted to give a final volume of 3.00 liters. the molarity of the final solution is ______.\n\n23) what is a substance called that can act as either an acid or a base. give an example of this kind of substance.\n\n24) what is the final concentration of a solution prepared by diluting 35.0 ml of 12.0 m hcl to a final volume of 1.20 l?\n\n25) what is the concentration of sodium chloride in the final solution if 25.34 ml of 0.113 m bacl₂ completely reacts and the total volume of the reaction is 110.4 ml, given the reaction: bacl₂ (aq) + na₂so₄(aq) → baso₄ (s) + 2nacl (aq)

show your work: (4 points each)\n19) a solution contains 100.0 g water, 10.0 g nacl, and 15.0 g methanol. what is the weight percent (w/w%) of methanol in the solution?\n\n20) what is the molarity of a solution prepared by dissolving 10.7 g nai in 0.250 l?\n\n21) determine the mass of cacl₂ (mm = 110.98) in 1.75 l of 1.50 m solution\n\n22) a 90.0 g sample of naoh is dissolved in water and the solution is diluted to give a final volume of 3.00 liters. the molarity of the final solution is ______.\n\n23) what is a substance called that can act as either an acid or a base. give an example of this kind of substance.\n\n24) what is the final concentration of a solution prepared by diluting 35.0 ml of 12.0 m hcl to a final volume of 1.20 l?\n\n25) what is the concentration of sodium chloride in the final solution if 25.34 ml of 0.113 m bacl₂ completely reacts and the total volume of the reaction is 110.4 ml, given the reaction: bacl₂ (aq) + na₂so₄(aq) → baso₄ (s) + 2nacl (aq)

Answer

19

Explanation:

Step1: Calculate total mass of solution

Total mass = mass of water+mass of NaCl + mass of methanol $m_{total}=100.0 + 10.0+15.0=125.0$ g

Step2: Calculate weight - percent of methanol

Weight - percent formula: $w/w%=\frac{mass\ of\ solute}{mass\ of\ solution}\times100%$ $w/w%=\frac{15.0}{125.0}\times 100% = 12.0%$

Answer:

$12.0%$

20

Explanation:

Step1: Calculate moles of NaI

Molar mass of NaI ($M$) = 149.89 g/mol. Moles ($n$) formula: $n=\frac{m}{M}$ $n=\frac{10.7}{149.89}\approx0.0714$ mol

Step2: Calculate molarity

Molarity ($M$) formula: $M=\frac{n}{V}$ (where $V$ is volume in liters) $M=\frac{0.0714}{0.250}=0.286$ M

Answer:

$0.286$ M

21

Explanation:

Step1: Calculate moles of $CaCl_2$

Molarity formula: $M = \frac{n}{V}$, so $n = M\times V$ $n=1.50\times1.75 = 2.625$ mol

Step2: Calculate mass of $CaCl_2$

Mass ($m$) formula: $m=n\times MM$ (where $MM$ is molar - mass) $m = 2.625\times110.98\approx291$ g

Answer:

$291$ g

22

Explanation:

Step1: Calculate moles of NaOH

Molar mass of NaOH ($MM$)=40.00 g/mol. Moles ($n$) formula: $n=\frac{m}{MM}$ $n=\frac{90.0}{40.00}=2.25$ mol

Step2: Calculate molarity

Molarity ($M$) formula: $M=\frac{n}{V}$ $M=\frac{2.25}{3.00}=0.750$ M

Answer:

$0.750$ M

23

Brief Explanations:

A substance that can act as either an acid or a base is called an amphoteric substance. Water ($H_2O$) is a common example as it can donate a proton (act as an acid) or accept a proton (act as a base).

Answer:

Amphoteric substance; $H_2O$

24

Explanation:

Step1: Use the dilution formula $M_1V_1 = M_2V_2$

Here, $M_1 = 12.0$ M, $V_1=35.0$ mL = 0.0350 L, $V_2 = 1.20$ L We need to find $M_2$. Rearranging the formula gives $M_2=\frac{M_1V_1}{V_2}$

Step2: Calculate $M_2$

$M_2=\frac{12.0\times0.0350}{1.20}=0.350$ M

Answer:

$0.350$ M

25

Explanation:

Step1: Calculate moles of $BaCl_2$

Molarity formula: $n = M\times V$ $n_{BaCl_2}=0.113\times0.02534 = 0.00286342$ mol

Step2: Determine moles of NaCl from the stoichiometry

From the reaction $BaCl_2(aq)+Na_2SO_4(aq)\to BaSO_4(s)+2NaCl(aq)$, the mole - ratio of $BaCl_2$ to $NaCl$ is 1:2. So $n_{NaCl}=2\times n_{BaCl_2}$ $n_{NaCl}=2\times0.00286342 = 0.00572684$ mol

Step3: Calculate molarity of NaCl

Molarity formula: $M=\frac{n}{V}$ (where $V = 0.1104$ L) $M=\frac{0.00572684}{0.1104}\approx0.0519$ M

Answer:

$0.0519$ M