Question #32915

1.61 grams of an hydrated salt BxH2O gave 0.714 grams of the an hydrous salt when heated. What will be the value of x if the relative molecular mass of B is 142?

Expert's answer

An hydrous salt is B.

If m(B) = 0.714 grams

M(B) = 142 g/mol

So, n(B) = m/M

n(B) = 0.714 g / 142g/mol = 0.005 mol;

We see, that

n(B) : n(BxH_{2}O) = 1 : 1;

So, n(BxH_{2}O) = 0.005 mol;

m(BxH_{2}O) = 1.61 grams

By the formula

M = m/n

we can find the molecular mass of an hydrated salt

M(BxH_{2}O) = 1.61 g / 0.005 mol = 322 g/mol

M(xH_{2}O) = M(BxH_{2}O) - M(B);

M(xH_{2}O) = 322 g/mol - 142 g/mol = 180 g/mol

if M(H_{2}O) = 18 g/mol

x = M(xH_{2}O) / M(H_{2}O)

x = 180 / 18 = 10

Answer: x = 10

If m(B) = 0.714 grams

M(B) = 142 g/mol

So, n(B) = m/M

n(B) = 0.714 g / 142g/mol = 0.005 mol;

We see, that

n(B) : n(BxH

So, n(BxH

m(BxH

By the formula

M = m/n

we can find the molecular mass of an hydrated salt

M(BxH

M(xH

M(xH

if M(H

x = M(xH

x = 180 / 18 = 10

Answer: x = 10

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