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Alex Ar [27]
3 years ago
10

3. A student adds 0.400g of n-propanol to 13.0 g of t-butanol. What is the molality of the solution? Show your calculations. (3

pts
Chemistry
1 answer:
marissa [1.9K]3 years ago
8 0

Answer:

m=0.512m

Explanation:

Hello,

In this case, we can consider the n-propanol as the solute (lower amount) and the t-butanol as the solvent (higher amount), for which, initially, we must compute the moles of n-propanol (molar mass = 60.1 g/mol) as shown below:

n_{solute}=0.400g*\frac{1mol}{60.1g0}=6.656x10^{-3}mol

Since the molality is computed via:

m=\frac{n_{solute}}{m_{solvent}}

Whereas the mass of the solvent is used in kilograms (0.0130g for the given one). Thus, we compute the resulting molality of the solution:

m=\frac{6.656x10^{-3}mol}{0.0130kg}\\ \\m=0.512\frac{mol}{kg}

Or just:

m=0.512m

Best regards.

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