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LuckyWell [14K]
3 years ago
10

In the laboratory you are asked to make a 0.565 m sodium bromide solution using 315 grams of water. How many grams of sodium bro

mide should you add? grams.
Chemistry
1 answer:
Scorpion4ik [409]3 years ago
6 0

Answer : The mass of sodium bromide added should be, 18.3 grams.

Explanation :

Molality : It is defined as the number of moles of solute present in kilograms of solvent.

Formula used :

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}

Solute is, NaBr and solvent is, water.

Given:

Molality of NaBr = 0.565 mol/kg

Molar mass of NaBr = 103 g/mole

Mass of water = 315 g

Now put all the given values in the above formula, we get:

0.565mol/kg=\frac{\text{Mass of NaBr}\times 1000}{103g/mole\times 315g}

\text{Mass of NaBr}=18.3g

Thus, the mass of sodium bromide added should be, 18.3 grams.

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What is the molar mass of a substance?
Sveta_85 [38]

Answer:

The mass in grams of one mole of a substance.

3 0
2 years ago
Determine the volume occupied by 0.352 mole of a gas at 25⁰C if the pressure is 81.8 kPa.
lions [1.4K]
Data:
p (pressure) = 81.8 kPa = 81.8*10³ Pa ≈ 8.07 atm
v (volume) = ? (in L)
n (number of mols) = 0.352 mol
R (Gas constant) = 0.082 (atm*L/mol*K)
T (temperature) = 25ºC converting to Kelvin, we have:
TK = TC + 273 → TK = 25 + 273 → TK = 298

Formula:
p*V =n*R*T

Solving:
p*V =n*R*T
8.07*V = 0.352*0.082*298
8.07V \approx 8.60
V \approx  \frac{8.60}{8.07}
\boxed{\boxed{V \approx 1.06\:L}}
7 0
3 years ago
Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown c
Alexus [3.1K]

Answer:

0.3229 M HBr(aq)

0.08436M H₂SO₄(aq)

Explanation:

<em>Stu Dent has finished his titration, and he comes to you for help with the calculations. He tells you that 20.00 mL of unknown concentration HBr(aq) required 18.45 mL of 0.3500 M NaOH(aq) to neutralize it, to the point where thymol blue indicator changed from pale yellow to very pale blue. Calculate the concentration (molarity) of Stu's HBr(aq) sample.</em>

<em />

Let's consider the balanced equation for the reaction between HBr(aq) and NaOH(aq).

NaOH(aq) + HBr(aq) ⇄ NaBr(aq) + H₂O(l)

When the neutralization is complete, all the HBr present reacts with NaOH in a 1:1 molar ratio.

18.45 \times 10^{-3} L NaOH.\frac{0.3500molNaOH}{1LNaOH} .\frac{1molHBr}{1molNaOH} .\frac{1}{20.00 \times 10^{-3} LHBr} =\frac{0.3229molHBr}{1LHBr} =0.3229M

<em>Kemmi Major also does a titration. She measures 25.00 mL of unknown concentration H₂SO₄(aq) and titrates it with 0.1000 M NaOH(aq). When she has added 42.18 mL of the base, her phenolphthalein indicator turns light pink. What is the concentration (molarity) of Kemmi's H₂SO₄(aq) sample?</em>

<em />

Let's consider the balanced equation for the reaction between H₂SO₄(aq) and NaOH(aq).

2 NaOH(aq) + H₂SO₄(aq) ⇄ Na₂SO₄(aq) + 2 H₂O(l)

When the neutralization is complete, all the H₂SO₄ present reacts with NaOH in a 1:2 molar ratio.

42.18 \times 10^{-3} LNaOH.\frac{0.1000molNaOH}{1LNaOH} .\frac{1molH_{2}SO_{4}}{2molNaOH} .\frac{1}{25.00\times 10^{-3}LH_{2}SO_{4}} =\frac{0.08436molH_{2}SO_{4}}{1LH_{2}SO_{4}} =0.08436M

6 0
3 years ago
How many grams of NaCL are required to prepare 50ml of a 2.0 molar solution​
Ksivusya [100]

Answer:

5.85 gm.

Explanation:

We know that,

Normality =<u> Molarity × Molecular </u><u>weight</u>

Equivalent weight

Since molecular weight of NaCl= equivalent weight = 23+35.5 =58.5

Normality of NaCl= molarity=2

Now,

Normality= <u>weight</u><u> </u><u>in</u><u> </u><u>gram</u><u> </u><u>×</u><u>1000</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>

Volume ×equivalent weight

Weight in gram is given by,

<u>=</u><u>Normality × Volume × equivalent </u><u>weight</u>

1000

= <u>2× 50 × 58.</u><u>5</u>

1000

=5.85 gm.

4 0
3 years ago
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