Question

1. calculate the molarity of 0.060 moles nahco₃ in 1500. ml of solution.

1. calculate the molarity of 29.25 grams of nacl in 2.0 liters of solution.

1. calculate the molarity of 34 grams of sugar, c₁₂h₂₂o₁₁ in 500. ml of solution.

1. calculate the number of moles of nacl contained in 0.500 l of a 1.5 m solution.

1. how many grams of nacl are contained in the solution discussed in problem #4?

Explanation:

Problem 1

Step1: Recall Molarity Formula

Molarity ($M$) is defined as moles of solute ($n$) divided by volume of solution in liters ($V$), so $M = \frac{n}{V}$.

Step2: Convert Volume to Liters

The volume is 1500 mL. Since 1 L = 1000 mL, $V = \frac{1500}{1000} = 1.5$ L.

Step3: Calculate Molarity

Given $n = 0.060$ mol and $V = 1.5$ L, substitute into the formula: $M = \frac{0.060}{1.5} = 0.040$ M.

Step1: Find Molar Mass of NaCl

Molar mass of Na is 22.99 g/mol and Cl is 35.45 g/mol, so molar mass of NaCl is $22.99 + 35.45 = 58.44$ g/mol.

Step2: Calculate Moles of NaCl

Moles ($n$) = mass ($m$) / molar mass ($M_m$), so $n = \frac{29.25}{58.44} \approx 0.5$ mol.

Step3: Calculate Molarity

Using $M = \frac{n}{V}$, with $n = 0.5$ mol and $V = 2.0$ L, $M = \frac{0.5}{2.0} = 0.25$ M.

Step1: Find Molar Mass of $\ce{C_{12}H_{22}O_{11}}$

Molar mass of C: 12.01 g/mol, H: 1.008 g/mol, O: 16.00 g/mol.
Molar mass = $12\times12.01 + 22\times1.008 + 11\times16.00 = 144.12 + 22.176 + 176.0 = 342.296$ g/mol.

Step2: Calculate Moles of Sugar

$n = \frac{m}{M_m} = \frac{34}{342.296} \approx 0.0993$ mol.

Step3: Convert Volume to Liters

500 mL = 0.5 L.

Step4: Calculate Molarity

$M = \frac{n}{V} = \frac{0.0993}{0.5} \approx 0.199$ M (or ~0.20 M).

Answer:

0.040 M

Problem 2