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kotegsom [21]
2 years ago
13

What mass of NaNO3 must be dissolved to make 838mL of a 1.25 M solution

Chemistry
1 answer:
denis-greek [22]2 years ago
4 0

Answer:

89.04 g of NaNO₃.

Explanation:

We'll begin by converting 838 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

838 mL = 838 mL × 1 L / 1000 mL

838 mL = 0.838 L

Next, we shall determine the number of mole of NaNO₃ in the solution. This can be obtained as follow:

Volume = 0.838 L

Molarity = 1.25 M

Mole of NaNO₃ =?

Mole = Molarity × volume

Mole of NaNO₃ = 1.25 × 0.838

Mole of NaNO₃ = 1.0475 mole

Finally, we shall determine the mass of NaNO₃ needed to prepare the solution. This can be obtained as follow:

Mole of NaNO₃ = 1.0475 mole

Molar mass of NaNO₃ = 23 + 14 + (16×3)

= 23 + 14 + 48

= 85 g/mol

Mass of NaNO₃ =?

Mass = mole × molar mass

Mass of NaNO₃ = 1.0475 × 85

Mass of NaNO₃ = 89.04 g

Therefore, 89.04 g of NaNO₃ is needed to prepare the solution.

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A quantity of gas occupies a volume of 804 mL at a temperature of 27 °C. At what temperature will the
irga5000 [103]

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T₂ = 150 K

Explanation:

Given data:

Initial volume of gas = 804 mL

Initial temperature = 27°C (27+273=300 K)

Final temperature = ?

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Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

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T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂ = V₂T₁/V₁  

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