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3 years ago

A venti latte (502mL) contains 0.322g caffeine, C8H10N4O2, express this concentration in molarity.

Chemistry

1 answer:

Natalija [7]3 years ago

7 0

Answer:

0.00331M

Explanation:

Molarity of a solution can be calculated as follows:

Molarity = number of moles (n) ÷ volume (V)

In the information provided in this question, the volume = 502mL = 502/1000 = 0.502L, mass of caffeine = 0.322g

Molar mass of caffeine (C8H10N4O2), where C = 12, H = 1, N = 14, O = 16 is as follows:

Molar mass = 12(8) + 1(10) + 14(4) + 16(2)

= 96 + 10 + 56 + 32

= 194g/mol

Using, moles = mass/molar mass

moles = 0.322/194

moles = 1.66 × 10-³ moles

mole = 0.00166mol

Molarity = number of moles ÷ volume

Molarity = 0.00166mol ÷ 0.502L

Molarity = 0.00331M.

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A 5.00 cm^3 sample of gold has a mass of 96.5 g. Calculate the density of gold.

Sholpan [36]

Answer:

$\boxed {\boxed {\sf 19.3 \ g/cm^3}}$

Explanation:

We are asked to calculate the density of gold. The density of a substance is its mass per unit volume. It is calculated using the following formula.

$\rho= \frac{m}{v}$

The mass of the sample is 96.5 grams. The volume of the sample is 5.00 cubic centimeters.

m= 96.5 g
v= 5.00 cm³

Substitute the values into the formula.

$\rho= \frac{ 96.5 \ g}{5.00 \ cm^3}$

Divide.

$\rho= 19.3 \ g/cm^3$

The density of gold is <u>19.3 grams per cubic centimeter.</u>

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Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of water is pro

Lubov Fominskaja [6]

This is an incomplete question, here is a complete question.

Aqueous sulfuric acid (H₂SO₄) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium sulfate (Na₂SO₄) and liquid water (H₂O) . If 12.5 g of water is produced from the reaction of 72.6 g of sulfuric acid and 77.0 g of sodium hydroxide, calculate the percent yield of water. Round your answer in significant figures.

Answer: The percent yield of water is, 46.8 %

Explanation : Given,

Mass of $H_2SO_4$ = 72.6 g

Mass of $NaOH$ = 77.0 g

Molar mass of $H_2SO_4$ = 98 g/mol

Molar mass of $NaOH$ = 40 g/mol

First we have to calculate the moles of $H_2SO_4$ and $NaOH$ .

$\text{Moles of }H_2SO_4=\frac{\text{Given mass }H_2SO_4}{\text{Molar mass }H_2SO_4}$

$\text{Moles of }H_2SO_4=\frac{72.6g}{98g/mol}=0.741mol$

and,

$\text{Moles of }NaOH=\frac{\text{Given mass }NaOH}{\text{Molar mass }NaOH}$

$\text{Moles of }NaOH=\frac{77.0g}{40g/mol}=1.925mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

From the balanced reaction we conclude that

As, 1 mole of $H_2SO_4$ react with 2 mole of $NaOH$

So, 0.741 moles of $H_2SO_4$ react with $0.741\times 2=1.482$ moles of $NaOH$

From this we conclude that, $NaOH$ is an excess reagent because the given moles are greater than the required moles and $H_2SO_4$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2O$