The molecular weight of an organic compound was determined by measuring the freezing point depression of a benzene solution. A 0.500 g sample was dissolved in 50.0 g of benzene, and the resulting depression was 0.42°C (kf (benzene) = 5.065°C/m). What is the approximate molecular weight of the compound?

The molar mass of the sample is 120.6 g/mol

Explanation:

To calculate the depression in freezing point, we use the equation:

Or,

where,

= depression in freezing point = 0.42°C

i = Vant hoff factor = 1 (For non-electrolytes)

= molal freezing point elevation constant = 5.065°C/m

= Given mass of solute (sample) = 0.500 g

= Molar mass of solute (sample) = ? g/mol

= Mass of solvent (benzene) = 50.0 g

Putting values in above equation, we get:

Hence, the molar mass of the sample is 120.6 g/mol

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation:

plasma is the state of matter where the substance is ionized and has no definite shape or volume. the fact of being ionized makes the substance a good electrical conductor.