--%>

Molality of glucose

Help me to go through this problem. Molecular weight of glucose is 180. A solution of glucose which contains 18 gms per litre is : (a) 2 molal (b) 1 molal (c) 0.1 molal (d)18 molal

   Related Questions in Chemistry

  • Q : Units of Measurement Unit of

      Unit of measurement- These are also some systems for units:      (1) 

  • Q : Finding Normality Can someone please

    Can someone please help me in getting through this problem. Concentrated H2SO4 has a density of 1.98 gm/ml and is 98% H2SO4 by weight. The normality is: (a) 2 N  (b) 19.8 N  (c) 39.6 N  (d) 98

  • Q : What type of bond does HCl encompass

    What type of bond does HCl encompass? Describe briefly?

  • Q : Group Cations Explain how dissolving

    Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid, establishes a buffer with a pH of approximately

  • Q : How to establish nomenclature for

    In the common chemistry terminologies, aliphatic halogen derivatives are named as alkyl halides. The words, n-, sec-, tert-, iso-, neo-, and amyl are

  • Q : Neutralization of sodium hydroxide How

    How much of NaOH is needed to neutralise 1500 cm3 of 0.1N HCl (given = At. wt. of Na =23): (i) 4 g  (ii) 6 g (iii) 40 g  (iv) 60 g

  • Q : Benefits of soapy detergents over the

    What are the benefits of soapy detergents over the soap less detergents? Briefly state the benefits?

  • Q : Organic and inorganic substances living

    living beings are made up of organic and inorganic substances.according to their complexity of their molecules how can ach of these substances be classified?

  • Q : Finding Active mass of urea Can someone

    Can someone please help me in getting through this problem. 10 litre solution of urea comprises of 240 gm urea. The active mass of urea is: (i) 0.04 (ii) 0.02 (iii) 0.4 (iv) 0.2

  • Q : Problem on Osmotic Pressure of solution

    The osmotic pressure of a 5% solution of cane sugar at 150oC  is (mol. wt. of cane sugar = 342)(a) 4 atm (b) 3.4 atm (c) 5.07 atm (d) 2.45 atmAnswer: (c) Π = (5 x 0.0821 x 1000 x 423)/(342 x 100) = 5.07 atm