JEE Main Solved Paper 2017 Question 28
Question: 1 gram of a carbonate $ (M _2CO _3) $ on treatment with excess HCl produces 0.01186 mole of $ CO _2. $ the molar mass of $ {M_2}C{O_3} $ in $ g,mo{l^{-1}} $ is:- JEE Main Solved Paper-2017
Options:
A) 1186
B) 84.3
C) 118.6
D) 11.86
Show Answer
Answer:
Correct Answer: B
Solution:
- Given chemical $ e{q^{n}} $ $ \underset{1,gm}{\mathop{M _2CO _3}},+2HCl\xrightarrow{{}}2MCl+H _2O+\underset{0.01186,mol}{\mathop{CO _2}}, $
$ \Rightarrow $ from the balanced chemical $ eq^{n}. $ $ \frac{1}{M}=0.01186 $
$ \Rightarrow $ $ ,gm/mol $ Molar mass of $M _2CO _3$=Mass of M _2CO _3 No. of moles of $M _2CO _3$=$\frac{1g}{0.01186mol}$=84.3g/mol
