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Free_Kalibri [48]
3 years ago
12

What mass of powdered drink mix is needed to make a 0.5 M solution of 100 mL?

Chemistry
2 answers:
son4ous [18]3 years ago
8 0

<u>Answer:</u> The mass of powdered drink mix is 17.1 grams.

<u>Explanation:</u>

The powdered drink mix has sucrose compound in it. So, we need to calculate the mass of sucrose in the given molar solution.

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of sucrose}\times \text{Volume of solution (in mL)}}

Molarity of solution = 0.5 M

Volume of solution = 100 mL

Molar mass of sucrose = 342.3 g/mol

Putting values in above equation, we get:

0.5M=\frac{\text{Mass of sucrose}\times 1000}{342.3g/mol\times 100mL}\\\\\text{Moles of sucrose}=\frac{0.5\times 342.3\times 100}{1000}=17.1g

Hence, the mass of powdered drink mix is 17.1 grams.

Maru [420]3 years ago
3 0

Answer:

There is 17,114825 g of powdered drink mix needed

Explanation:

<u>Step 1 :</u> Calculate moles

As given, the concentration of the drink is 0.5 M, this means 0.5 mol / L

Since the volume is 100mL, we have to convert the concentration,

⇒0.5 / 1   =  x /0.1    ⇒ 0.5* 0.1  = x = 0.05 M

This means there is 0.05 mol per 100mL

e

<u>Step 2 </u>: calculate mass of the powdered drink

here we use the formula n (mole) = m(mass) / M (Molar mass)

⇒ since powdered drink mix is usually made of sucrose (C12H22O11) and has a molar mass of 342.2965 g/mol.

0.05 mol = mass / 342.2965 g/mol

To find the mass, we isolate it ⇒0.05 mol * 342.2965 g/mol = 17,114825g

There is 17,114825 g of powdered drink mix needed

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Answer:

V₂  =279.9 cm³

Explanation:

Given data:

Initial volume = 360 cm³

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Solution:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Solution:

<em>We will convert the mmHg to atm.</em>

700/760 = 0.92 atm

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50+273 = 323 K

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V₂ = 0.92 atm × 360 cm³ × 273 K / 323 K ×1 atm

V₂  =  290417.6 atm .cm³.  K  / 323 k. atm

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3 years ago
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What effect does increasing the concentration of a dissolved solute have on each of the colligative properties?
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Answer:

If we increase the concentration of a dissolved solute, the solution would have a vapor pressure so much low, the boiling temperature for the solution will be so high, freezing point for the solution will be so much low and the osmotic pressure will be higher.

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