chemical-aritmatic

Question: Ratio of ${C_p}$and ${C_v}$of a gas X is 1.4, the number of atom of the gas ‘X’ present in 11.2 litres of it at NTP will be

1) $6.02 \times {10^{23}}$

2) $1.2 \times {10^{23}}$

3) $3.01 \times {10^{23}}$

4) $2.01 \times {10^{23}}$

Solution:

Explanation: No Explanation

atomic-mass

Rate this question:

★ ★ ★ ★ ★

Average rating: (0 votes)