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3 years ago

the mass by mass percentage of a solution is 20%. find the mass of the solute dissolved in 500g of solution

Chemistry

1 answer:

GuDViN [60]3 years ago

6 0

Answer:

$m_{solute}=100g$

Explanation:

Hello,

In this case, by considering the mathematical definition of by mass percentage of a solute as shown below:

$\%m/m=\frac{m_{solute}}{m_{solution}}*100\%$

We are able to compute the mass of the solute in a 20% solution having 500 g of solution as follows:

$m_{solute}=\frac{m_{solution}*\%m/m}{100\%} =\frac{500g*20\%}{100\%}\\ \\m_{solute}=100g$

Best regards.

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Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . Suppose 6.9 g

satela [25.4K]

Answer:

$5.6gNa_2SO_4$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2NaOH(aq)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)$

Therefore, since the masses of both of the reactants are given, one computes the available moles of sulfuric acid and those moles of it consumed by the sodium hydroxide as shown below:

$n_{H_2SO_4}^{available}=6.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.0704molH_2SO_4\\n_{H_2SO_4}^{consumed\ by\ NaOH}=3.14gNaOH*\frac{1molNaOH}{40gNaOH}*\frac{1molH_2SO_4}{2molNaOH}=0.04molH_2SO_4$

In such a way, since there is more available sulfuric acid than it that is consumed, the sodium hydroxide is the limiting reagent, consequently, the maximum mass of sodium sulfate turns out:

$m_{Na_2SO_4}=0.04molH_2SO_4*\frac{1molNa_2SO_4}{1molH_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4}=5.6gNa_2SO_4$

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5 0

2 years ago

6. What causes the phases of the moon?

lawyer [7]

Answer:

The rotation of the Earth.

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2 years ago

Four samples of gas each exert 675 mm Hg in separate 3.5 L containers. What pressure will they exert if they are all placed in a

DedPeter [7]

The pressure that will be exerted if four sample of gas are placed in a single 3.5 container is calculated as below

if each gas occupies 675 mmhg

what about 4 gases in the sample
by cross multiplication

= 675 mm hg x 4/1 = 2.7 x10^3mmhg (answer D)

3 0

3 years ago

What is the predicted change in the boiling point of water when 1.50 g of

dezoksy [38]

Answer:

0.00735°C

Explanation:

By seeing the question, we can see the elevation in boiling point with addition of BaCl₂ in water

⠀

$\textsf {While} \: \sf {\Delta T_b} \: \textsf{expression is used} \\ \textsf {for elevation of boiling point}$

⠀

The elevation in boiling point is a phenomenon in which there is increase in boiling point in solution, when the particular type of solute is added to pure solvent.

⠀

$\sf \large \underline{The \: formula \: to \: be \: used \: in \: this \: question \: is} \\ \boxed{T_b = i \times K_b \times m}$

⠀

Where 'i' is van't hoff factor which represents the ratio of observed osmotic pressure and the value to be expected.

and 'i' is 3 (as given in the question)

⠀

'Kb' is molal boiling point constant. And it's value is 0.51°C/mol(given in question)

⠀

'm' represent the molality of solution. Molatity is no. of moles of solution present in 1kg of solution.

⠀

To find molality, we have to divide no. of moles of solute by weight of solution

⠀

While first we need to no. of moles

$\sf \implies no. \: of \: moles = \frac{weight \: of \: solute}{molar \: mass \: of \: solute} \\ \\ \implies \sf no. \: of \: moles = \frac{1.5}{208.23} \\ \\ \sf \implies no. \: of \: moles = 0.0072$

⠀

Now, we will find molality

⠀

$\sf \hookrightarrow molality = \frac{no.\: of \: moles}{weight \: of \: solution} \\ \\ \sf \hookrightarrow molality = \frac{0.072}{1.5} \\ \\ \sf \hookrightarrow molality = 0.048 \: mol {kg}^{ - 1}$

⠀

$\textsf{ \large{ \underline{Now substituting the required values}}}$

⠀

$\sf \longmapsto \Delta T_b = 3 \times 0.51 \times 0.0048 \\ \\ \\ \boxed{ \tt{ \longmapsto \Delta T_b =0.00735{ \degree}C}}$

⠀

Henceforth, the change in boiling point is 0.00735°C.

7 0

1 year ago

Calculate the heat required to convert 5.0 g of ice at 0.0 c to steam at 100

Fiesta28 [93]

Heat= latent heat of fusion+sensible heat+ latent heat of vapourization
=(79.7*5)+(5*100*1)+(540*5)
=3598.5 cal

6 0

3 years ago

the mass by mass percentage of a solution is 20%. find the mass of the solute dissolved in 500g of solution​

the mass by mass percentage of a solution is 20%. find the mass of the solute dissolved in 500g of solution