JEE Main 2017 · 2 Apr · Shift 1 · Chemistry Q32

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ChemistrySingle correct+4 / -1Question 32 of 90

1 gram of a carbonate ( $\ce{M2CO3}$ ) on treatment with excess HCl produces 0.01186 mole of $\ce{CO2}$ . The molar mass of $\ce{M2CO3}$ in g mol $^{-1}$ is :

118.6

11.86

1186

84.3

Answer and worked solution

Official answer: D

One mole of $\ce{M2CO3}$ gives one mole of $\ce{CO2}$ , so $M=\dfrac1{0.01186}=84.3$ g mol⁻¹.

Topics:Mole concept

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JEE Main 2017 · 2 Apr · Shift 1, Chemistry question 32