How many grams of solid NaCl (Formula Weight: 58.44 g/mole) are required to prep

Question

How many grams of solid NaCl (Formula Weight: 58.44 g/mole) are required to prepare 150 mL of a 250mM solution? If you dissolve 15mg glucose (Formula Weight: 180.16g/mole) in 10mL of water (all answers must be presented in scientific notation format): What is the concentration of glucose in Mm and mu M? If you take 5ml of this solution and dilute it to 120ml what is the concentration of the glucose in the resulting solution in mu M? How many nanomoles of glucose are present in 10mL of the diluted solution?

Explanation / Answer

moles of NaCL in 150ml of 250 mM= Molarity* Volume in L= 250*10-3 Moles/L* (150/1000)L= 0.0375 moles

since moles= mass/molar mass, mass of NaCl required= moles* molar mass= 0.0375*58.5 gm =2.19 gm

2. mass of glucose =15 mg =15*10-3 gm, moles of glucose = mass/molar mass =15*10-3/180 =8.33*10-5 moles

This is dissolved in 10 ml of water, volume of water in L= 10/1000=0.01L

Concentration of glucose in M= 8.33*10-5/0.01=0.00833 M

when converted into mM, this becomes 0.00833*1000 mM ( since 1000mM= 1M)

concentration of glucose= 8.33 mM

in uM, 0.00833*106uM= 8.33*103 uM

since M1V1= M2V2. M1= molarity of solution of 0.00833M V1= volume of water= 5ml, V2= 120 ml,

5*0.00833= M2*120

M2= 0.000347 M or 0.000347*106 uM=347 uM

10 ml of this solution contains 0.000347M*10/1000 moles of Glucose =3.47*10-6 moles

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