The volume of 0.1 M HCl required in mL to n | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given,Molarity of $HCl$ $(M_1) = 0.1 \ M$ Volume of $Na_2CO_3$ solution $(V_2) = 20 \ mL$ Mass of $Na_2CO_3 = 0.106 \ g$ Molar mass of $Na_2CO_3 =

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(C) Given,Molarity of $HCl$ $(M_1) = 0.1 \ M$ Volume of $Na_2CO_3$ solution $(V_2) = 20 \ mL$ Mass of $Na_2CO_3 = 0.106 \ g$ Molar mass of $Na_2CO_3 = (2 	imes 23) + 12 + (3 	imes 16) = 106 \ g/mol$ Moles of $Na_2CO_3 = \frac{0.106 \ g}{106 \ g/mol} = 0.001 \ mol$ Molarity of $Na_2CO_3$ $(M_2) = \frac{0.001 \ mol}{0.020 \ L} = 0.05 \ M$ The neutralization reaction is: $Na_2CO_3 + 2HCl ightarrow 2NaCl + H_2O + CO_2$ Using the equivalence relation: $n_{HCl} = 2 	imes n_{Na_2CO_3}$ $M_1 	imes V_1 = 2 	imes (M_2 	imes V_2)$ $0.1 	imes V_1 = 2 	imes (0.05 	imes 20)$ $0.1 	imes V_1 = 2 	imes 1 = 2$ $V_1 = \frac{2}{0.1} = 20 \ mL$

The volume of $0.1 \ M \ HCl$ required in $mL$ to neutralise $20 \ mL$ of a solution containing $0.106 \ g$ of $Na_2CO_3$ is

Similar Questions

$x \ g$ of methane was burnt completely in the presence of oxygen. The liberated gases were passed into a solution containing $370 \ g$ of $Ca(OH)_2$. The weight of white precipitate obtained was $500 \ g$. What is the value of $x$ (in $g$)? (Given: $C=12, H=1, Ca=40, O=16 \ u$)

The molarity of a $0.2 \ N \ Na_2CO_3$ solution is ........ $M$.

......... $cm^3$ is the volume of $0.1 \ N \ HCl$ required to react completely with $1.0 \ g$ of pure calcium carbonate $(Ca = 40, C = 12, O = 16)$.

At constant temperature and pressure,the change in volume when $80\%\, 100\, mL$ of phosgene $(COCl_2)$ decomposes into $CO$ and $Cl_2$ is an increase of ............... $mL$.

In the combustion of $2.0 \ g$ of methane,$25 \ kcal$ of heat is liberated. The heat of combustion of methane would be ..... $kcal$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module