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3 years ago

The normal freezing point of a certain liquid

Chemistry

1 answer:

slavikrds [6]3 years ago

3 0

Answer : The molal freezing point depression constant of liquid X is, $4.12^oC/m$

Explanation : Given,

Mass of urea (solute) = 5.90 g

Mass of liquid X (solvent) = 450 g = 0.450 kg

Molar mass of urea = 60 g/mole

Formula used :

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of urea}}{\text{Molar mass of urea}\times \text{Mass of liquid X Kg}}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution = $-0.5^oC$

$\Delta T^o$ = freezing point of liquid X = $0.4^oC$

i = Van't Hoff factor = 1 (for non-electrolyte)

$K_f$ = Molal-freezing-point-depression constant = ?

m = molality

Now put all the given values in this formula, we get

$0.4^oC-(-0.5^oC)=1\times K_f\times \frac{5.90g}{60g/mol\times 0.450kg}$

$K_f=4.12^oC/m$

Therefore, the molal freezing point depression constant of liquid X is, $4.12^oC/m$

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3 years ago

barium chloride (Ba(CIO3)2) breaks down to form barium chloride and oxygen. what is the balanced equation for this equation?

kompoz [17]

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3 years ago

1.20 x 10^22 molecules NaOH to gram

Trava [24]

Answer:

$\boxed {\boxed {\sf 0.797 \ g \ NaOH}}$

Explanation:

<u>1. Convert Molecules to Moles</u>

First, we must convert molecules to moles using Avogadro's Number: 6.022*10²³. This tells us the number of particles in 1 mole of a substance. In this case, the particles are molecules of sodium hydroxide.

$\frac {6.022*10^{23} \ molecules \ NaOH} {1 \ mol \ NaOH}}$

Multiply by the given number of molecules.

$1.20*10^{22} \ molecules \ NaOH *\frac {6.022*10^{23} \ molecules \ NaOH} {1 \ mol \ NaOH}}$

Flip the fraction so the molecules cancel out.

$1.20*10^{22} \ molecules \ NaOH *\frac {1 \ mol \ NaOH} {6.022*10^{23} \ molecules \ NaOH}}$

$1.20*10^{22} *\frac {1 \ mol \ NaOH} {6.022*10^{23}}}$

$\frac {1.20*10^{22} \ mol \ NaOH} {6.022*10^{23}}}$

$0.0199269345732 \ mol \ NaOH$

<u>2. Convert Moles to Grams</u>

Next, we convert moles to grams using the molar mass.

We must calculate the molar mass using the values on the Periodic Table. Look up each individual element.

Na: 22.9897693 g/mol
O: 15.999 g/mol
H: 1.008 g/mol

Since the formula has no subscripts, we can simply add the molar masses.

NaOH: 22.9897693+15.999+1.008=39.9967693 g/mol

Use this as a ratio.

$\frac {39.9967693 \ g \ NaOH }{1 \ mol \ NaOH}$

Multiply by the number of moles we calculated.

$0.0199269345732 \ mol \ NaOH*\frac {39.9967693 \ g \ NaOH }{1 \ mol \ NaOH}$

The moles of sodium hydroxide cancel.

$0.0199269345732 *\frac {39.9967693 \ g \ NaOH }{1}$

$0.0199269345732 *39.9967693 \ g \ NaOH$

$0.79701300498 \ g \ NaOH$

The original measurement of molecules has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place. The 0 tells us to leave the 7 in the hundredth place.

$0.797 \ g \ NaOH$