SATHEE: Some Basic Concepts of Chemistry

Some Basic Concepts of Chemistry - Result Question 19

####19. Specific volume of cylindrical virus particle is $6.02 \times 10^{- 2} c c / g$ whose radius and length are $7 \AA$ $Misplaced &$ respectively. If $N_{A} = 6.02 \times 10^{23}$ , find molecular weight of virus

[2001]

(a) $3.08 \times 10^{3} k g / m o l$

(b) $3.08 \times 10^{4} k g / m o l$

(d) $15.4 k g / m o l$

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Solution:

(d) Specific volume (volume of $1 g$ ) of cylindrical virus particle $= 6.02 \times 10^{- 2} c c / g$

Radius of virus $(r) = 7 \AA = 7 \times 10^{- 8} c m$

Length of virus $= 10 \times 10^{- 8} c m$

Volume of virus $=$

$π r^{2} l = \frac{22}{7} \times (7 \times 10^{- 8})^{2} \times 10 \times 10^{- 8}$

$= 154 \times 10^{- 23} c c$

Wt. of one virus particle $= \frac{volume}{specific volume}$

Mol. wt. of virus $=$ Wt. of $N_{A}$ particle

$= \frac{154 \times 10^{- 23}}{6.02 \times 10^{- 2}} \times 6.02 \times 10^{23}$

$= 15400 g / m o l = 15.4 k g / m o l$

Some Basic Concepts of Chemistry - Result Question 19

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Some Basic Concepts of Chemistry - Result Question 19

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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