The vapor pressure of water is 2376 mm hg at 25c how many...

The vapor pressure of water is 23.76 mm Hg at 25°C. How many grams of glucose, C6H12O6, a nonvolatile, nonelectrolyte (MW = 180.2 g/mol), must be added to 250.3 grams of water to reduce the vapor pressure to 23.18 mm Hg ? water = H2O = 18.02 g/mol. g glucose

Transcript

00:01 In this question, vapor pressure of water is given to us.

00:03 So the, simply the vapor pressure of water equals to 23 .76 millimeters of hg at 25 degrees celsius.

00:15 Now the questions ask that how many grams of glucose? so we have to calculate the amount of glucose in this amount should be in gram, that it should be added as a non -volatile non -electrolite.

00:30 Compound to the 250 grams of water so the amount of water h2 is given to us is to 50 .3 gram of water so that the vapor pressure becomes reduced that is vapor pressure of this solution now becomes 23 .18 m m of hg so this is what we have to do now in this case we will be using a rauts law and according to the rhodes law the mole fraction in this case let's say the mole fraction of glucose equals to the vapor pressure of this pure solvent and that is h2o minus the vapor pressure of a solution that is after addition of glucose this whole and divided by the vapor pressure of h2 so this is how we calculate or define the route slow so for this particular solution first of all we need in the number of moles of this h2o so the number of moles of h2o is simply that in the given amount that is 250 .3 gram whole divided by the molecular mass of water molecule that is 18 gram and it comes out to be 13 .89 moles of h2 okay so now we have the number of most of h2 and the amount of glucose is basically is to be calculated in this case.

02:07 So the mole fraction of glucose becomes the number of moles of glucose divided by number of moles of glucose plus the number of moles, number of moles of water.

02:23 And the value for this vapor pressure of h2 is given to us as 23 .76 minus 23 .18 whole divided by 23 .76...

Question

The vapor pressure of water is 23.76 mm Hg at 25°C. How many grams of glucose, C6H12O6, a nonvolatile, nonelectrolyte (MW = 180.2 g/mol), must be added to 250.3 grams of water to reduce the vapor pressure to 23.18 mm Hg ? water = H2O = 18.02 g/mol. g glucose

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