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MissTica
2 years ago
13

5. What is the mole fraction of NaOH in an aqueous solution that contains 31.0 % NaOH by mass?

Chemistry
1 answer:
Liono4ka [1.6K]2 years ago
6 0

Answer:

16.8%      

Explanation:

31% NaOH    molar mass  40 gm

69% H2O      molar mass  18  gm

1000 gm would be  

 310 gm NaOH  or  310/40 = 7.75 moles

 690 gm of H2O or  690/18 = 38.333 moles

7.75 / (7.75 + 38.333) = .168   mole fraction

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Read 2 more answers
A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?
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Answer:

At -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

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Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

                                          \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}

where P_{1} and P_{2} are initial and final pressure respectively.

           V_{1}  and V_{2} are initial and final volume respectively.

           T_{1} and T_{2} are initial and final temperature in kelvin scale respectively.

Here P_{1}=150.0kPa , V_{1}=1.75L , T_{1}=(273-23)K=250K, P_{2}=210.0kPa and V_{2}=1.30L

Hence    T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}

            \Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}

            \Rightarrow T_{2}=260K

            \Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}

So at -13 ^{0}\textrm{C} , the gas would occupy 1.30L at 210.0 kPa.

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3 years ago
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