# Question 3dbe1

Jan 18, 2018

$a . \approx 183 m l$
$b . \approx 260 g$

#### Explanation:

A. Assuming that the concentration of the solution presented is in %w/w; thus the formula shown below is applicable to solve this problem.

%solution(w/w)=("mass solute"(m))/("mass solution"(m_s))
where:

% solution =17%
$\text{mass solute} = 35 g N a C l$
$\rho \text{ solution} = \frac{1.13 g}{c {m}^{3}}$

Now, plug in given data to its respective variable to get the intermediate value needed to solve the prescribed problem. Rearrange formula as needed to isolate the required variable; that is,

%solution(w/w)=m/m_s
m_s=(m)/(%solution)
${m}_{s} = \frac{35 g}{0.17}$
${m}_{s} = 205.8824 g$

Knowing the ${m}_{s}$, the total volume of the solution that also contained the given mass of the solute can be calculated through the density formula; thus,

${\rho}_{s} = {m}_{s} / {V}_{s}$
where:

${\rho}_{s} = \text{density of the solution}$
${m}_{s} = \text{mass of the solution}$
${V}_{s} = \text{volume of the solution}$

${V}_{s} = {m}_{s} / {\rho}_{s}$
${V}_{s} = \frac{205.8824 \cancel{g}}{\frac{1.13 \cancel{g}}{m l}}$
${V}_{s} \approx 183 m l$

B. Assuming that the concentration of the solution presented is again in %w/w; thus the formula shown below is applicable to solve this problem.

%solution(w/w)=("mass solute"(m))/("mass solution"(m_s))
where:

% solution =58%
$\text{mass solute} = 150 g {H}_{2} S {O}_{4}$

Now, plug in given data to its respective variable to get the intermediate value needed to solve the prescribed problem. Rearrange formula as needed to isolate the required variable; that is,

%solution(w/w)=m/m_s
m_s=(m)/(%solution)#
${m}_{s} = \frac{150 g}{0.58}$
${m}_{s} = 260 g$