Home
Class 11
CHEMISTRY
4.215 g a metallic carbonate was heated ...

4.215 g a metallic carbonate was heated in a hard glass tube and `CO_2` evolved was found to measure 1336 mL at 27°C and 700 mm pressure. What is the equivalent mass of the metal ?

Text Solution

Verified by Experts

The correct Answer is:
A, B

`Pv = nRT = (W)/(Mw) RT`
`:. (700)/(760) xx (1336)/(1000) = (W)/(44) xx 0.0821 xx 300`
or `W = 2.198 g CO_(2)`
The molecular mass of any metal carbonate is `2E + 60`,
Where `E` is the equivalent mass of the metal. So
`(2E + 60)/(4.215) = (44)/(2.193)`
`E = 12.188 g`
Promotional Banner

Topper's Solved these Questions

  • SOME BASIC CONCEPTS AND MOLE CONCEPT

    CENGAGE CHEMISTRY ENGLISH|Exercise Archives Fill In The Blanks|4 Videos
  • S-BLOCK GROUP 2 - ALKALINE EARTH METALS

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex 5.1 Objective|2 Videos
  • STATES OF MATTER

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises (Ture False)|25 Videos

Similar Questions

Explore conceptually related problems

4.2 g of a metallic carbonate MCO_(3) was heated in a hard glass tube and CO_(2) evolved was found to have 1120 mL of volume at STP . The Ew of the metal is

240 mL of a dry gas measured at 27°C and 750 mm pressure weighed 0.64 g. What is the molecular mass of the gas ?

When dissolved in dilute sulphuric acid, 0.275 g of a metal evolved 119.7 mL of hydrogen at 20°C and 763 mm pressure. What is the equivalent mass of the metal ?

0.05 g of magnesium when treated with dilute HCI gave 51 mL of hydrogen at 27°C and 780 mm pressure. Calculate the equivalent mass of magnesium.

0.30 g of gas was found to occupy a volume of 82.0 mL at 27^(@)C and 3 atm. Pressure. The molecular mass of the gas is

A given mass of a gas occupies 143 cm^3 at 27^@C and 700 mm Hg pressure. What will be its volume at 300 K and 280 mm Hg pressure?

In Duma's method 0.52g of an organic compound on combustion gave 68.6 mL N_(2) at 27^(@)C and 756mm pressure. What is the percentage of nitrogen in the compound?

1.725 g of a metal carbonate is mixed with 300 mL of N/10 HCI. 10 mL of N/2 sodium hydroxide were required to neutralise excess of the acid. Calculate the equivalent mass of the metal carbonate.

1.50 g of a metal on being heated in oxygen gives 2.15 g of its oxide. Calculate the equivalent mass of the metal.

10 g of a metal carbonate on heating given 5.6 g of its oxide. The equivalent amount metal 5x. What is x.