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3241004551 [841]
3 years ago
7

If 120 g of NaOH is dissolved in 2.5 L of water, what is the molarity of the solution?

Chemistry
1 answer:
klio [65]3 years ago
8 0

Answer:

1.2 M.

Explanation:

<em>Molarity is the no. of moles of solute in a 1.0 L of the solution.</em>

M = n/V.

<em>M = (mass/molar mass)solute x (1000/V of the solution).</em>

mass of NaOH = 120.0 g.

molar mass of NaOH = 40.0 g/mol.

V of the solution = 2.5 L = 2500.0 mL.

<em>∴ M = (mass/molar mass)solute x (1000/V of the solution)</em> = (120.0 g / 40.0 g/mol) x (1000 / 2500.0 mL) = <em>1.2 M.</em>

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emmasim [6.3K]
When it comes to equilibrium reactions, it useful to do ICE analysis. ICE stands for Initial-Change-Equilibrium. You subtract the initial and change to determine the equilibrium amounts which is the basis for Kc. Kc is the equilibrium constant of concentration which is just the ratio of products to reactant. 

Let's do the ICE analysis

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I         0        1.3    1.65
C     +2x       -x      -3x
-------------------------------------
E       0.1        ?        ?

The variable x is the amount of moles of the substances that reacted. You apply the stoichiometric coefficients by multiplying it by x. Now, we can solve x by:

Equilibrium NH₃ = 0.1 = 0 + 2x
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K_{C} = \frac{[ C^{c} ]}{[ A^{a} ][ B^{b} ]}

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3 years ago
A certain substance melts at a temperature of . But if a sample of is prepared with of urea dissolved in it, the sample is found
pshichka [43]

Answer:

2.2 °C/m

Explanation:

It seems the question is incomplete. However, this problem has been found in a web search, with values as follow:

" A certain substance X melts at a temperature of -9.9 °C. But if a 350 g sample of X is prepared with 31.8 g of urea (CH₄N₂O) dissolved in it, the sample is found to have a melting point of -13.2°C instead. Calculate the molal freezing point depression constant of X. Round your answer to 2 significant digits. "

So we use the formula for <em>freezing point depression</em>:

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In this case, ΔTf = 13.2 - 9.9 = 3.3°C

m is the molality (moles solute/kg solvent)

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Molality = 0.53 / 0.35 = 1.51 m

So now we have all the required data to <u>solve for Kf</u>:

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<span>4p orbital</span>

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