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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
States of Matter

NTA Abhyas 2020 · Chemistry · States of Matter

The van der Waal's constant 'b' of a gas is 4 π × 1 0 ^ - 4 L/mol ....

The van der Waal's constant 'b' of a gas is 4π×10−4L/mol4 \pi \times 1 {0}^{- 4} L/mol. How near can the centres of the two molecules approach each other ? [Use : NA=6×1023{N}_{A} = 6 \times {10}^{23}]

Options

  1. A10−7m1 {0}^{- 7} m
  2. B10−10m1 {0}^{- 10} m
  3. C5×10−11m5 × 1 {0}^{- 1 1} m
  4. D5×10−9m5 × 1 {0}^{- 9} m

Correct answer

B. 10−10m1 {0}^{- 10} m

Step-by-step solution

b=4×43×πr3×NA\text{b} = 4 \times \dfrac{4}{3} \times \pi r^{3} \times {N}_{A} ;

We know that, 1L=1000cm31L = 1000{ cm}^{3}

4×π×10−4×1000=4×43×π×r3×6×10234 \times \pi \times 1 {0}^{- 4} \times 1 0 0 0 = 4 \times \dfrac{4}{3} \times \pi \times r^{3} \times 6 \times 1 {0}^{2 3}

r=5×10−9cm\text{r} = 5 \times 1 {0}^{- 9} cm

Distance of closest approach =2r= 2r

=10−8cm= 1 {0}^{- 8} cm or 10−10m1 {0}^{- 10} m
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