Home
Class 12
CHEMISTRY
1 gram of a carbonate (M(2)CO(3)) o trea...

1 gram of a carbonate `(M_(2)CO_(3))` o treatment with excess HCl produces 0.01186 mole of `CO_(2)`. The molar mass of `M_(2)CO_(3)` in g `mol^(-1)` is:-

A

`1186`

B

`84.3`

C

`118.6`

D

`11.86`

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Topper's Solved these Questions

  • MOLE CONCEPT

    ALLEN|Exercise J-ADVANCE|2 Videos
  • MOLE CONCEPT

    ALLEN|Exercise O-II (Match the column:)|3 Videos
  • METALLURGY

    ALLEN|Exercise EXERCISE-05 [B]|26 Videos
  • p-Block Element

    ALLEN|Exercise All Questions|20 Videos

Similar Questions

Explore conceptually related problems

1g of a carbonate (M_(2)CO_(3)) on treatment with excess HCl produces 0.01186 mole of CO_(2) . The molar mass of M_(2)CO_(3) in g mol^(-1) is

1g of a carbonate (M_(2)CO_(3)) on treatment with excess HCl produces 0.01186 mole of CO_(2) . The molar mass of M_(2)CO_(3) in g mol^(-1) is

The volume strength of 1 M H_(2)O_(2) is : (Molar mass of H_(2)O_(2) = 34 g mol^(-1)

The volume strength of 2M H_(2)O_(2) is (Molar mass of H_(2)O_(2) = 34 g mol^(-1) )

The volume strength of g1 M H_(2)O_(2) is : (Molar mass of H_(2)O_(2) = 34 g mol^(-1) )

What is the mass of 0.1 mole of CO_(2) in gram?

62.5 gm of a mixture of CaCO_(3) and SiO_(2) are treated with excess of HCl and 1.1 gm of CO_(2) is produced. What is mass % CaCO_(3) in the mixture.

70 g of a sample of magnesite on treatment with excess of HCl gave 11.2 L of CO_(2) at STP. The percentage purify of the sample

CaCO_(3) + 2HCl to CaCl_(2) + H_(2)O + CO_(2) 0.2 mole CaCO_(3) is reacted with excess of HCl. Find mass of CO_(2) produced?

How are 0.50 mol Na_(2)CO_(3) and 0.50 M Na_(2)CO_(3) different?