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Answer :
We begin by finding the volume of the solution. The total mass of the solution is the sum of the mass of the carbohydrate and water:
[tex]$$
\text{Mass}_{\text{total}} = 6.37\,\text{g} + 100.2\,\text{g} = 106.57\,\text{g}.
$$[/tex]
Since the density of the solution is given as
[tex]$$
\rho = 1.024\,\frac{\text{g}}{\text{mL}},
$$[/tex]
the volume in milliliters is
[tex]$$
\text{Volume (mL)} = \frac{\text{Mass}_{\text{total}}}{\rho} = \frac{106.57\,\text{g}}{1.024\,\frac{\text{g}}{\text{mL}}} \approx 104.07\,\text{mL}.
$$[/tex]
Converting to liters (since the gas constant is given in L atm/(mol K)):
[tex]$$
V = \frac{104.07\,\text{mL}}{1000} \approx 0.10407\,\text{L}.
$$[/tex]
Next, convert the temperature from Celsius to Kelvin:
[tex]$$
T = 20.0\,^\circ\text{C} + 273.15 = 293.15\,\text{K}.
$$[/tex]
The osmotic pressure is related to the molar concentration (number of moles, [tex]$n$[/tex], in volume [tex]$V$[/tex]) by the formula
[tex]$$
\pi = \frac{n}{V}RT,
$$[/tex]
where [tex]$R = 0.0821\,\frac{\text{L atm}}{\text{mol K}}$[/tex]. Solving for [tex]$n$[/tex], we get
[tex]$$
n = \frac{\pi\,V}{R\,T}.
$$[/tex]
Substitute the known values:
[tex]$$
n = \frac{(4.61\,\text{atm})(0.10407\,\text{L})}{(0.0821\,\frac{\text{L atm}}{\text{mol K}})(293.15\,\text{K})} \approx 0.01993\,\text{mol}.
$$[/tex]
Now, the molar mass [tex]$M$[/tex] is given by the mass of the solute divided by the number of moles:
[tex]$$
M = \frac{\text{mass of carbohydrate}}{n} = \frac{6.37\,\text{g}}{0.01993\,\text{mol}} \approx 320\,\frac{\text{g}}{\text{mol}}.
$$[/tex]
Thus, the molar mass of the carbohydrate is
[tex]$$
\boxed{320\,\frac{\text{g}}{\text{mol}}}.
$$[/tex]
[tex]$$
\text{Mass}_{\text{total}} = 6.37\,\text{g} + 100.2\,\text{g} = 106.57\,\text{g}.
$$[/tex]
Since the density of the solution is given as
[tex]$$
\rho = 1.024\,\frac{\text{g}}{\text{mL}},
$$[/tex]
the volume in milliliters is
[tex]$$
\text{Volume (mL)} = \frac{\text{Mass}_{\text{total}}}{\rho} = \frac{106.57\,\text{g}}{1.024\,\frac{\text{g}}{\text{mL}}} \approx 104.07\,\text{mL}.
$$[/tex]
Converting to liters (since the gas constant is given in L atm/(mol K)):
[tex]$$
V = \frac{104.07\,\text{mL}}{1000} \approx 0.10407\,\text{L}.
$$[/tex]
Next, convert the temperature from Celsius to Kelvin:
[tex]$$
T = 20.0\,^\circ\text{C} + 273.15 = 293.15\,\text{K}.
$$[/tex]
The osmotic pressure is related to the molar concentration (number of moles, [tex]$n$[/tex], in volume [tex]$V$[/tex]) by the formula
[tex]$$
\pi = \frac{n}{V}RT,
$$[/tex]
where [tex]$R = 0.0821\,\frac{\text{L atm}}{\text{mol K}}$[/tex]. Solving for [tex]$n$[/tex], we get
[tex]$$
n = \frac{\pi\,V}{R\,T}.
$$[/tex]
Substitute the known values:
[tex]$$
n = \frac{(4.61\,\text{atm})(0.10407\,\text{L})}{(0.0821\,\frac{\text{L atm}}{\text{mol K}})(293.15\,\text{K})} \approx 0.01993\,\text{mol}.
$$[/tex]
Now, the molar mass [tex]$M$[/tex] is given by the mass of the solute divided by the number of moles:
[tex]$$
M = \frac{\text{mass of carbohydrate}}{n} = \frac{6.37\,\text{g}}{0.01993\,\text{mol}} \approx 320\,\frac{\text{g}}{\text{mol}}.
$$[/tex]
Thus, the molar mass of the carbohydrate is
[tex]$$
\boxed{320\,\frac{\text{g}}{\text{mol}}}.
$$[/tex]
Thank you for reading the article A solution of 6 37 g of a carbohydrate in 100 2 g of water has a density of tex 1 024 frac g mL. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
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Rewritten by : Jeany
