Question

A commercial sample of oxalic acid is labeled as 22.5% by w/w. calculate (a) molarity (b) volume of acid required to prepare 1l of 0.2 m oxalic acid, h2c2o4.

Answer 1

please give the correct question it is incorrect question I think so

Answer 2

Answer: a) 3.75 M

b) 0.05 L

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n= moles of solute

 $V_s$ = volume of solution in ml

Given : 22.5 g of oxalic acid dissolved in 100 g of the solution.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{22.5}{90}=0.25moles$

Density = 1.5 g/ml

Thus volume of solution=$\frac{\text {mass of solution}}{\text{density of solution}}=\frac{100}{1.5}=66.6ml$

$Molarity=\frac{0.25\times 1000}{66.6}=3.75M$

b) The expression used will be :

$M_1V_1=M_2V_2$

where,

$M_1$ = concentration of stock oxalic acid= 3.75 M

$M_2$ = concentration of resulting oxalic acid = 0.2 M

$V_1$ = volume of stock oxalic acid = ? L

$V_2$ = volume of resulting oxalic acid  =1 L

$3.75\times V_1=0.2\times 1L$

By solving the terms, we get :

$V_1=0.05L$

Therefore, the volume of the acid required to prepare 1l of 0.2 m oxalic acid, will be 0.05 L