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fomenos
4 years ago
15

A solution is prepared by dissolving 50.4 gg sucrose (C12H22O11C12H22O11) in 0.332 kgkg of water. The final volume of the soluti

on is 355 mLmL. For this solution, calculate the molarity.
Chemistry
1 answer:
elena-s [515]4 years ago
4 0

Answer:

The correct answer is 0.41 M

Explanation:

The molarity of a solution is given by the number of moles of solute per liter of solution.

Molarity= moles solute/1 L solution= mol/L

First we need the molecular weight (Mw) of sucrose in order to convert the mass of sucrose (in g) to mol. We can quickly calculate this from the molar mass (MM) of the elements C, H and O in Periodic Table:

Mw sucrose (C₁₂H₂₂O₁₁)= (12 x MM C) +(22 x MM H) + (11 x MM O)= (12 x 12 g/mol)+(22 x 1 g/mol) + (11 x 16 g/mol)= 342 g/mol

Then, we divide the mass of sucrose (50.4 g) into the molecular weight of sucrose to obtain the number of moles and this is divided into the volume of the solution (355 ml). We convert ml to L taking into account that 1000 ml= 1L:

50.4 g sucrose x 1 mol/342 g sucrose x 1/355 ml x 1000 ml/1 L= 0.415 mol/L= 0.41 M

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it is c. ionic compound just because it says ionic. can't be a crystal since there's no links to an ionic bond connected to crystals (and if it were it'd depend on the elements and the thingy majig itself right) and a covalent bond is an entirely different bond which isn't related to ionic. hope this helped, and i hope you have a good day too! :D

3 0
3 years ago
Cu+2AgNO
baherus [9]

Answer:

Mass of Ag produced = 64.6 g

Note: the question is, how many grams of Ag is produced from 19.0 g of Cu and 125 g of AgNO3

Explanation:

Equation of the reaction:

Cu + 2AgNO3 ---> 2Ag + Cu(NO3)2

From the equation above, 1 mole of Cu reacts with 2 moles of AgNO3 to produce 2 moles of Ag and 1 mole of Cu(NO3)2.

Molar mass of the reactants and products are; Cu = 63.5 g/mol, Ag = 108 g/mol, AgNO3 = 170 g/mol, Cu(NO3)2 = 187.5 g/mol

To determine, the limiting reactant;

63.5 g of Cu reacts with 170 * 2 g of AgNO3,

19 g of Cu will react with (340 * 19)/63.5 g of AgNO3 =101.7 g of AgNO3.

Since there are 125 g of AgNO3 available for reaction, it is in excess and Cu is the limiting reactant.

63.5 g of Cu reacts to produce 108 * 2 g of Ag,

19 g of Cu will react to produce (216 * 19)/63.5 g of Ag = 64.6 g of Ag.

Therefore mass of Ag produced = 64.6g

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