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Ilya [14]
2 years ago
9

Someone help pls thank you i will give brainliest

Chemistry
2 answers:
Yanka [14]2 years ago
7 0

It's the same as that other guy answered:

(Plus, I'm not plagiarizing, I'm just answering that it's the same answer as the other guy commented.)

Lelu [443]2 years ago
6 0

Answer: See attached image

Explanation:

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At 2°C, the vapor pressure of pure water is 23.76 mmHg and that of a certain seawater sample is 23.09 mmHg. Assuming that seawat
wariber [46]

Answer:

0.808  M

Explanation:

Using Raoult's Law

\frac{P_s}{Pi}= x_i

where:

P_s = vapor pressure of sea water( solution) = 23.09 mmHg

P_i = vapor pressure of pure water (solute) = 23.76 mmHg

x_i = mole fraction of water

∴

\frac{23.09}{23.76}= x_i

x_i = 0.9718

x_i+x_2=1

x_2 = 1- x_i

x_2 = 1- 0.9718

x_2 = 0.0282

x_i = \frac{n_i}{n_i+n_2}  ------ equation (1)

x_2 = \frac{n_2}{n_i+n_2}  ------ equation (2)

where; (n_2) = number of moles of sea water

(n_i) = number of moles of pure water

equating above equation 1 and 2; we have :

\frac{n_2}{n_i}= \frac{0.0282}{0.9178}

= 0.02901

NOW, Molarity =  \frac{moles of sea water}{mass of pure water }*1000

= \frac{0.02901}{18}*1000

= 0.001616*1000

= 1.616 M

As we assume that the sea water contains only NaCl, if NaCl dissociates to Na⁺ and Cl⁻; we have \frac{1.616}{2} =0.808 M

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2 years ago
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