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3 years ago

What would the total mass of a 4.9 mole sample of Barium (Ba) be if its Molar Mass is 137g?

Chemistry

1 answer:

Aleksandr-060686 [28]3 years ago

8 0

Answer:

m = 671 grams

Explanation:

Given that,

No of moles, n = 4.9

Molar mass of Barium, M = 137 g

Mass divided by molar mass is equal to no of moles. It can be given by the formula as follows :

$n=\dfrac{m}{M}\\\\m=n\times M\\\\m=4.9\times 137\\\\m=671.3\ g$

m = 671 grams

So, the total mass of the sample of Barium is 671 grams.

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4 years ago

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

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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.

⠀

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

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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.

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2 years ago