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Rama09 [41]
3 years ago
9

Calculate the percent by mass of 4.35g of Na I dissolved in 105g of water​

Chemistry
1 answer:
Ludmilka [50]3 years ago
4 0

<u>We are given:</u>

Mass of Na added = 4.35 grams

Mass of water = 105 grams

<u>Mass Percent of Na:</u>

Total mass of the solution = mass of solute + mass of solvent

Total mass of the solution = 4.35 + 105 = 109.35 grams

Mass percent of solute = (mass of solute / mass of solution) * 100

Mass percent of Solute = (4.35 / 109.35) * 100

Mass percent = 3.978 %

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A mixture containing 20 mole % butane, 35 mole % pentane and rest
notka56 [123]

Answer:

2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB

Whereas y accounts for the fractions at the outlet distillate and x for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

0.9=\frac{y_bD}{z_bF}

So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

y_bD=0.9*z_bF=0.9*0.2*100mol=18mol

The total distillate flow:

y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol

And the total bottoms flow:

F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025

Then, the molar fraction of pentane and hexane:

x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422

x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

Regards.

8 0
3 years ago
Read 2 more answers
What is the solution's freezing point: 15 g of CH4N2O (Molar mass = 60.055 g/mol) in 200. g of H2O? (Kf = 1.86 (°C·kg)/mol)
AURORKA [14]

Ok to answer this question we firsst need to fin the number of mol of Urea (CH4N2O). to do this we simply :

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therefore 200g of water contain 0.25mol

the next step is to determine the malality of our solution in 200g of water, to do this we say:

200 g = 1Kg/1000g = 0.2kg

therefor 0.25mol/0.2Kg = 1.25mol/kg

and from the equation:

we know that i = 1

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therefore;

the solutions freezing point is -2.325°C

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