How many mL of a 0.250M sodium hydroxide solution is needed to neutralize 25.0 mL of a 0.430M sulfuric acid solution?
1 answer:
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Answer : The molar mass of the solute will be 87.90 g/mol.
Explanation : We know the formula for elevation in boiling point, which is
Δt = i
m
given that, Δt = 0.357, = 5.02 and mass of 
= 40,
on substituting the value we get,
0.357 = (1) X (5.02) X (x/ 0.044), on solving we get x = 2.844 X
.
Now, 0.250/ 2.844 X = 87.90 g/mol. which is the weight of unknown component.
Explanation : We know the formula for elevation in boiling point, which is
Δt = i
given that, Δt = 0.357,
on substituting the value we get,
0.357 = (1) X (5.02) X (x/ 0.044), on solving we get x = 2.844 X
Now, 0.250/ 2.844 X