Answer: The molality of solution is 0.66 mole/kg
Explanation:
Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.
![Molarity=\frac{n\times 1000}{W_s}](https://tex.z-dn.net/?f=Molarity%3D%5Cfrac%7Bn%5Ctimes%201000%7D%7BW_s%7D)
where,
n = moles of solute
= weight of solvent in g
moles of
= ![\frac{\text {given mass}}{\text {Molar Mass}}=\frac{58.0g}{174g/mol}=0.33mol](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%20%7Bgiven%20mass%7D%7D%7B%5Ctext%20%7BMolar%20Mass%7D%7D%3D%5Cfrac%7B58.0g%7D%7B174g%2Fmol%7D%3D0.33mol)
Now put all the given values in the formula of molality, we get
![Molality=\frac{0.33\times 1000}{500g}=0.66mole/kg](https://tex.z-dn.net/?f=Molality%3D%5Cfrac%7B0.33%5Ctimes%201000%7D%7B500g%7D%3D0.66mole%2Fkg)
Therefore, the molality of solution is 0.66 mole/kg
B is the correct answer tbh
The mass of NaCl sample has been 24.3 g. Thus, option A is correct.
The heat of fusion has been the amount of heat required to convert 1 mole of substance into solid to liquid state.
The heat required has been given as:
![Q=m\Delta H](https://tex.z-dn.net/?f=Q%3Dm%5CDelta%20H)
<h3>Computation for the mass of NaCl</h3>
The given solution has heat of fusion, ![\Delta H_{fus}=30.2\rm kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H_%7Bfus%7D%3D30.2%5Crm%20kJ%2Fmol)
The heat required to melt the sample has been, ![Q=732.6\;\rm kJ/mol](https://tex.z-dn.net/?f=Q%3D732.6%5C%3B%5Crm%20kJ%2Fmol)
Substituting the values for the mass of NaCl
![\rm 732.6\;kJ/mol=Mass\;\times\;30.2\;kJ/mol\\\\ Mass=\dfrac{732.6}{30.2}\;g\\\\ Mass=24.3\;g\\](https://tex.z-dn.net/?f=%5Crm%20732.6%5C%3BkJ%2Fmol%3DMass%5C%3B%5Ctimes%5C%3B30.2%5C%3BkJ%2Fmol%5C%5C%5C%5C%0AMass%3D%5Cdfrac%7B732.6%7D%7B30.2%7D%5C%3Bg%5C%5C%5C%5C%0A%20Mass%3D24.3%5C%3Bg%5C%5C)
The mass of NaCl sample has been 24.3 g. Thus, option A is correct.
Learn more about heat of fusion, here:
brainly.com/question/87248
A gas made up of atoms escapes through a pinhole 0.225times as fast as gas. Write the chemical formula of the gas.
Answer:
Explanation:
To solve this problem, we must apply Graham's law of diffusion. This law states that "the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molecular mass at constant temperature and pressure".
Mathematically;
![\frac{r_{1} }{r_{2} } = \frac{\sqrt{m_{2} } }{\sqrt{m_{1} } }](https://tex.z-dn.net/?f=%5Cfrac%7Br_%7B1%7D%20%7D%7Br_%7B2%7D%20%7D%20%20%3D%20%5Cfrac%7B%5Csqrt%7Bm_%7B2%7D%20%7D%20%7D%7B%5Csqrt%7Bm_%7B1%7D%20%7D%20%7D)
r₁ is the rate of diffusion of gas 1
r₂ is the rate of diffusion of gas 2
m₁ is the molar mass of gas 1
m₂ is the molar mass of gas 2
let gas 2 be the given H₂;
molar mass of H₂ = 2 x 1 = 2gmol⁻¹
rate of diffusion is 0.225;
i .e r1/r2 = 0.225
0.225 = √2 / √ m₁
0.225 = 1.414 / √ m₁
√ m₁ = 6.3
m₁ = 6.3² = 39.5g/mol
The gas is likely Argon since argon has similar molecular mass