Question

calculate the volume required of 63% (by mass) solution of HNO3 in water to completely react with 100ml 5 molar solution of NaOH given density of HNO3 solution is 1.5g/cm3​

Answer 1

Answer: 33.3 ml

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 : 63 g of $HNO_3$ is present in 100 g of solution

moles of solute = $\frac{\text {given mass}}{\text {molar mass}}=\frac{63g}{63g/mol}=1mol$

Density of $HNO_3$ solution = $1.5g/cm^3$

Volume of $HNO_3$ solution =$\frac{mass}{density}=\frac{100g}{1.5g/cm^3}=66.7cm^3=66.7ml$  

 $(1cm^3=1ml)$

Now put all the given values in the formula of molarity, we get

$Molarity=\frac{1\times 1000}{66.7ml}$

$Molarity=15M$

According to the neutralization law,

$n_1M_1V_1=n_2M_2V_2$

where,

$M_1$ = molarity of NaOH solution = 5 M

$V_1$ = volume of NaOH solution = 100 ml

$M_2$ = molarity of $HNO_3$ solution = 15 M

$V_2$ = volume of $HNO_3$ solution = ?

$n_1$ = valency of $NaOH$ = 1

$n_2$ = valency of $HNO_3$ = 31

$1\times (5M)\times 100ml=1\times (15M)\times V_2$

$V_2=33.3ml$

Therefore,  the volume required of 63% (by mass) solution of $HNO_3$ in water to completely react with 100ml of 5 molar solution of $NaOH$ is 33.3 ml