Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol.
We know Molarity = moles of the solute / volume of solution
Putting the given values in above equation,we get
0.15 = (mole of benzoic acid / 250) x 1000
Or
moles of benzoic acid = (0.15 x 250) / 1000
= 0.0375 mol of benzoic acid
Also molecular mass of benzoic acid (C6H5COOH)
= 7 × 12 + 6 × 1 + 2 × 16
= 122 g/mol
Therefore amount of benzoic acid = 0.0375 x 122 = 4.575 g
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 30: Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in ....
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