Question

A mass of 18 gram of compound M2CO3. xH2O (molar mass of M = 24 g mol–1) is dissolved in 500 mL water and titrated against 0.5 M HCl solution. If the volume of HCl consumed is 500 mL, the value of x is

Answer 1

molarity=no. of mole of solute/volm of solution in liter*1000
             0.5=18/108x+18/(500*1000)
             by this calculate the value of x
hope i am correct

Answer 2

Answer: The value of x is 2.

Explanation: Normality of a solution is defined as the number of gram equivalents of solute dissolved per Liter of the solution.

$Normality=\frac{\text{given mass of compound}}{\text{Equivalent mass of compound}\times \text{volume of solution in L}}$

${\text {Equivalent mass}}=\frac{text {Molecular mass}}{Valency}$

$Normality=\frac{18g}{\frac{108+18x}{2}}\times 0.5L}}$

$Normality=\frac{18g}{54+9x}\times 0.5L}}$

Also According to the neutralization law,

$N_1V_1=N_2V_2$

where,

$N_1$ = Normality of $M_2CO_3.xH_2O$ solution =?

$V_1$ = volume of $M_2CO_3.xH_2O$ solution = 500 ml

$N_2$ = Normality of HCl solution = 0.5 M

$V_2$ = volume of HCl solution = 500 ml

Now put all the given values in the above law, we get the volume of NaOH solution.

$N_1\times 500ml=(0.5M)\times (500ml)$

$N_1=0.5N$

$0.5M=\frac{18g}{54+9x}\times 0.5L}}$

$x=2$