What is the molality of a d-mannose solution prepared by dissolving 45.0 g of d-mannose, c6h12o6, in 125.9 g of water?

The molality of 45 g of D- mannose in 125.9 g of water will be 1.985 m.

Explanation:

Molality is the number of moles of a solute in a solution per kg of solvent.

Molality =

Molecular weight of = 180.156 grams

Mass of water = 125.9 g = 0.1259 kg

Molality =

Molality = 1.985 m

The molality of a d-mannose solution prepared by dissolving 45 g of d-mannose in 125.9 g of water has the molality of 1.985 m.

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1. 1 mole of anything contains 6.022x10^23 pieces  

d  

2. a  

3. c, number  

5. 1.24x10^24atoms / 6.022x10^23atoms/mole = 2moles h2 atoms  

6. fe2br3 mass = 2moles x mass fe + 3 moles x mass br  

molar mass = grams / number of moles = g/mole  

to find moles = g / molar mass  

to find mass = g = molar mass / moles

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g