Discussion Forum



1)

1 g of a carbonate (M2CO3) on treatment with excess HCl produces 0.01186 mole of CO2 . The molar mass M2CO3 in g mol-1 is


A) 1186

B) 84.3

C) 118.6

D) 11.86

Answer:

Option B

Explanation:

$M_{2}CO_{3}+2HCl\rightarrow 2MCl+H_{2}O+CO_{2}$

1g                                                                  0.01186 mole

 

Number of moles of M2CO3 reacted = Number of moles of CO2 evolved

$\frac{1}{M}=$  0.01186 [M = Molar mass of M2CO3]

$M=\frac{1}{0.01186}=84.3 g mol^{-1}$