Q1 / 90 · JEE Main 2017 (02 Apr) · Chemistry

$1$ gram of a carbonate $(M_{2} {CO}_{3})$ on treatment with excess $HCl$ produces $0.01186$ moles of ${CO}_{2}$ . The molar mass of $M_{2} {CO}_{3}$ in ${gmol}^{- 1}$ is:

Correct answer
A
84.3
B
118.6
C
11.86
D
1186

Solution

$M_{2} C O_{3} + 2 H C l \to 2 M C l + H_{2} O + C O_{2}$

From reaction,

$Moles of M_{2} C O_{3} = M o l e s o f C O_{2}$

$\frac{Weight of M_{2} {CO}_{3}}{Molar Mass of M_{2} {CO}_{3}} = \frac{Weight of {CO}_{2}}{Molar mass of {CO}_{2}}$

$\frac{1}{Molar Mass of M_{2} {CO}_{3}} = \frac{0.01186 \times 44}{44}$

$\frac{1}{Molar Mass of M_{2} {CO}_{3}} = 0 .01186$

Molar Mass $= \frac{1}{0.01186} = 84.3$ g/mol

Hence, the correct answer is option: 'A'.

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1 gram of a carbonate M2CO3 on treatment with excess HCl produces 0.01186 moles of CO2 . The molar mass of M2CO3 in gmol-1 is:

Solution

Relevant Questions

$1$ gram of a carbonate $(M_{2} {CO}_{3})$ on treatment with excess $HCl$ produces $0.01186$ moles of ${CO}_{2}$ . The molar mass of $M_{2} {CO}_{3}$ in ${gmol}^{- 1}$ is: