The number of grams of oxygen in 32.2 g of N | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The atomic masses of elements are: $Na = 23 \ g/mol$, $S = 32 \ g/mol$, $O = 16 \ g/mol$, $H = 1 \ g/mol$. The molecular mass of $Na_2SO_4 \cdot 1

Vedclass · Accepted Answer

$22.4$

Vedclass · Answer

(B) The atomic masses of elements are: $Na = 23 \ g/mol$, $S = 32 \ g/mol$, $O = 16 \ g/mol$, $H = 1 \ g/mol$. The molecular mass of $Na_2SO_4 \cdot 10H_2O = (2 	imes 23) + 32 + (4 	imes 16) + (10 	imes 18) = 46 + 32 + 64 + 180 = 322 \ g/mol$. Given mass of $Na_2SO_4 \cdot 10H_2O = 32.2 \ g$. Moles of $Na_2SO_4 \cdot 10H_2O = \frac{32.2 \ g}{322 \ g/mol} = 0.1 \ mol$. In one mole of $Na_2SO_4 \cdot 10H_2O$, there are $4$ oxygen atoms in $Na_2SO_4$ and $10$ oxygen atoms in $10H_2O$, totaling $14$ oxygen atoms. Moles of oxygen atoms $= 0.1 \ mol 	imes 14 = 1.4 \ mol$. Mass of oxygen $= 1.4 \ mol 	imes 16 \ g/mol = 22.4 \ g$.

The number of grams of oxygen in $32.2 \ g$ of $Na_2SO_4 \cdot 10H_2O$ is approximately (in $g$)

Similar Questions

The formula of a substance is $(CO)_x$. If its vapor density is $70$,find the value of $x$.

An organic compound containing $C$, $H$, and $O$ has $13.13 \% H$, $52.14 \% C$, and $34.73 \% O$. Its molar mass is $46.068 \ g/mol$. What are its empirical and molecular formulae?

Compound $A$ contains $8.7\,\%$ Hydrogen,$74\,\%$ Carbon and $17.3\,\%$ Nitrogen. The molecular formula of the compound is ....
Given : Atomic masses of $C$,$H$ and $N$ are $12$,$1$ and $14\, amu$ respectively.
The molar mass of the compound $A$ is $162\, g\, mol^{-1}$.

$A$ compound gave $80\%$ carbon and $20\%$ hydrogen on analysis. The compound is possibly

Empirical formula of a compound is $CH_2O$ and its vapour density is $30$. The molecular formula of the compound is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module