Silver crystallises in fcc lattice. If edge length of the cell is 4.07 x 10-8cm and density is 10.5 g cm-3, calculate the atomic mass of silver.
From the equation, density = ZxM/a3 x N0, we get
M = ρ x a3 x N0/Z …………………………………1
Placing the values in equation 1,
WE GET M= 10.5 x (4.07 x10-8)3 x 6.022 x 1023/4
= 106.6 gmol-1
= Therefore, atomic mass of silver = 106.6 gmol-1
The rate constant for the decomposition of N2O5 at various temperatures is given below:
T/°C |
0 | 20 | 40 | 60 | 80 |
105 X K /S-1 |
0.0787 | 1.70 | 25.7 | 178 | 2140 |
Draw a graph between ln k and 1/T and calculate the values of A and Ea.
Predict the rate constant at 30 º and 50 ºC.
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Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 11: Silver crystallises in fcc lattice. If edge length of the cell is 4.07 x 10-8cm and density is 10.5 ....
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