Question

112 ml of No2 at stp was liquified , the density of liquid being 1.15 g/ml . Calculate the volume and the no of molecules of liquid No2

Answer 1

Answer:

Explanation:

Mass of NO2 = Molar mass of NO2/Volume of the solution in Lx

Volume of NO2 = 46/22400x112=0.23 g; Volume of the Liquid = Mass of the liquid / density = 0.23/1.15 = 0.20 ml; No of molecules in NO2 = moles x avogadro number = 0.23/46 x 6.022 x 10 to the power 23 = 3.01 x 10 to the power 23

Answer 2

Explanation:

At STP the moles any gas if Volume V (litres) is given by:

$n = \frac{V}{22.4}$

thus moles of NO2

n = 112 × 10^-3 L / 22.4 L

= 5 × 10^-3 mol

Now The mass of NO2 is calculated by the formula

$n = \frac{given \: \: mass}{molar \: mass}$

Thus mass = moles × molar mass

molar mass of NO2 = 46 g

given mass = 46 × 5 × 10^-3 = 2.3 × 10^-1 g

Now

$V \: = \frac{m}{d}$

V = 2.3 × 10^-1 / 1.15

= 2×10^-1

= 0.2 mL

molecules in 5 × 10^-3 mol = (5 × 10^-3) × Na

= ~ 3.01 × 10^21 molecules