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Allisa [31]
3 years ago
7

How many grams of sodium chloride are dissolved in 375 ml of a 0.90% saline solution?

Chemistry
1 answer:
liq [111]3 years ago
3 0

Answer:

337.5 g NaCl

Explanation:

.90% saline solution = 90g saline (NaCl)/100mL water (H2O)

1. Use 375 mL to find mass

375 mL × (90 g NaCl/100 mL) = 337.5 g NaCl in 375 mL .9% saline solution

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Air at 30.0% relative humidity is cooled isobarically at 1 atm absolute from 75.0°C to 35.0°C. a. Estimate the dew point and the
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Answer:

a. Dew point: 48.7°C. Degrees of superheat 26.3°C

b. 3.33mol/m^3

c. P=3.34atm

Explanation:

a. Based on the psychometric chart of air, the specific volume of air at the given conditions is:

v =\frac{RT}{PM} =\frac{0.082\frac{atm*L}{mol*K}*348.15K}{1atm*28.97g/mol} *\frac{1m^3}{1000L}*\frac{1000g}{1kg}  =0.98544m^3/kg

The dew point at the specific volume and the 30%-humidity has a value of 48.7°C, it means that there are 75°C-48.7°C=26.3°C of superheat.

b. At 75°C the molar fraction of water is 11580Pa/101625Pa=0.114 moles per cubic meter of feed gas are:

\frac{n}{V}=y_{H_2O}\frac{P}{RT}=0.114\frac{1atm}{0.082\frac{atm*L}{mol*K}*348.15K} }*\frac{1000L}{1m^3}    \\\frac{n}{V}=4mol/m^3

Once the 35°C are reached, the mole fraction of water is 1688Pa/101325Pa=0.017 and remaining moles per cubic meter of feed gas are:

\frac{n}{V}=y_{H_2O}\frac{P}{RT}=0.017\frac{1atm}{0.082\frac{atm*L}{mol*K}*308.15K} }*\frac{1000L}{1m^3}    \\\frac{n}{V}=0.673mol/m^3

So the condensed moles per cubic meter of feed gas are:

4mol/m^3-0.673mol/m^3=3.33mol/m^3

c. Considering the Raoult's law, one computes the pressure as follows:

P=\frac{P^{sat}_{H_2O}}{y_{H_2O}}

At 75°C and 30%-humidity, the saturation water vapor pressure has a value of 38599Pa, thus:

P=\frac{38599Pa}{0.114}*\frac{1atm}{101325Pa} \\P=3.34atm

Best regards.

6 0
4 years ago
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