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stealth61 [152]
3 years ago
10

Calculate molality,molarity and mole fraction of KI if the density of 20%(mass) aqueous KI is 1.202 g/mL

Chemistry
1 answer:
Scorpion4ik [409]3 years ago
5 0
Basis: 1 L of the substance.
               (1.202 g/mL) x (1000 mL) = 1202 g 
                   mass solute = (1202 g) x 0.2 = 240.2 g
                   mass solvent = 1202 g x 0.8 =  961.6 g
                   moles KI = (240.2 g) x (1 mole / 166 g)  = 1.45 moles
                   moles water = (961.6 g) x (1 mole / 18 g) = 53.42 moles
1. Molality = moles solute / kg solvent 
                  = 1.45 moles / 0.9616 kg = 1.5 m
2. Molarity = moles solute / L solution
                  = 1.45 moles / 1 L solution = 1.45 M
3. molar mass = mole solute / total moles
                        =  1.45 moles / (1.45 moles + 53.42 moles) = 0.0264 

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q = mCΔT

The correct specific heat capacity of water is <em>4.187 kJ/(kg.K)</em>.

ΔT = q/mC = 87 kJ/[648.00 kg x 4.187 kJ/(kg.K)] = 87 kJ/(2713 kJ/K) = 0.032 K

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A syringe initially holds a sample of gas with a volume of 285 mL at 355 K and 1.88 atm. To what temperature must the gas in the
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<u>Answer:</u> The temperature to which the gas in the syringe must be heated is 720.5 K

<u>Explanation:</u>

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1,V_1\text{ and }T_1 are the initial pressure, volume and temperature of the gas

P_2,V_2\text{ and }T_2 are the final pressure, volume and temperature of the gas

We are given:

P_1=1.88atm\\V_1=285mL\\T_1=355K\\P_2=2.50atm\\V_2=435mL\\T_2=?K

Putting values in above equation, we get:

\frac{1.88atm\times 285mL}{355K}=\frac{2.50atm\times 435mL}{T_2}\\\\T_2=\frac{2.50\times 435\times 355}{1.88\times 285}=720.5K

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PLS HELP QUICK ALOTTT OF POINTS
timofeeve [1]

Answer:

\boxed {\boxed {\sf 0.80 \ mol\ F}}

Explanation:

We are asked to find how many moles are in 4.8 × 10²³ fluorine atoms. We convert atoms to moles using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of fluorine.

We will convert using dimensional analysis and set up a ratio using Avogadro's Number.

\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}

We are converting 4.8 × 10²³ fluorine atoms to moles, so we multiply the ratio by this number.

4.8 \times 10^{23} \ atoms \ F *\frac {6.022 \times 10^{23} \ atoms \ F}{ 1 \ mol \ F}

Flip the ratio so the units of atoms of fluorine cancel each other out.

4.8 \times 10^{23} \ atoms \ F *\frac { 1 \ mol \ F}{6.022 \times 10^{23} \ atoms \ F}

4.8 \times 10^{23}  *\frac { 1 \ mol \ F}{6.022 \times 10^{23} }

Condense into 1 fraction.

\frac { 4.8 \times 10^{23} }{6.022 \times 10^{23} } \ mol \ F

Divide.

0.7970773829 \ mol \ F

The original measurement of atoms has 2 significant figures, so our answer must have the same. For the number we found, that is the hundredths place. The 7 in the thousandths tells us to round the 9 in the hundredths place up to a 0. Then, we also have to round the 7 in the tenths place up to an 8.

0.80 \ mol \ F

4.8 × 10²³ fluorine atoms are equal to <u>0.80 moles of fluorine.</u>

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